In what follows, for \(n=1,\,2\) , the notations
\(H \equiv \frac{\rho T}{\varrho },\qquad G \equiv \frac{5\mathcal {E}}{2T} - \frac{3\rho ^2 }{2\varrho },\qquad p_n\equiv \frac{\rho T}{\varrho }\rho _n+\frac{G}{3}\theta _n,\) 
are used for convenience. Then there holds
\(\partial _t G +\nabla \cdot (G\mathfrak {u})=0,\qquad (\partial _t + \mathfrak {u}\cdot \nabla )H +\frac{2}{3}H\nabla \cdot \mathfrak {u}=0.\) 
With the notations and identities above, the hyperbolic system (REF ) can be written as
\(\left\lbrace \begin{aligned}&\partial _t p_n + \mathfrak {u}\cdot \nabla _x p_n + \frac{5}{3}\mathcal {E}\nabla \cdot u_n + \frac{5}{3}(\nabla \cdot \mathfrak {u}) p_n + u_n\cdot \nabla \mathcal {E}=\frac{1}{3}\mathcal {F}_\theta ^\bot (F_n)\\&\rho \partial _t u_n + \rho \mathfrak {u}\cdot \nabla u_n +\nabla p_n - \frac{\nabla \mathcal {E}}{\rho ^2T/\varrho }p_n + \rho u_n\cdot \nabla \mathfrak {u}- \frac{\nabla \mathcal {E}}{3\rho ^2T/(G\varrho )}\theta _n = \mathcal {F}_u^\bot (F_n)\\& G \partial _t \theta _n +G \mathfrak {u}\cdot \nabla \theta _n +2GT\nabla \cdot u_n +3Gu_n\cdot \nabla T + \frac{2}{3}G (\nabla \cdot \mathfrak {u})\theta _n = \mathcal {F}_\theta ^\bot (F_n).\end{aligned}\right.\) 
Then a symmetric hyperbolic equation is obtained:
\(A_0^n \partial _t U_n + \sum _{j=1}^{3}A_j^n \partial _j U_n + A_4^n U_n = \mathcal {F}_n,\) 
where
\(U_n=\left(\begin{array}{c}p_n \\ u_n \\ \theta _n \end{array}\right),\quad A_0^n=\left(\begin{array}{ccc} \frac{9}{5} & 0 & -\frac{\mathcal {E}}{T} \\ 0 & \rho \mathcal {E}\,\mathbb {I} & 0\\ -\frac{\mathcal {E}}{T} & 0 & \frac{5\mathcal {E}^2}{6T^2} \end{array}\right),\quad A_j^n=\left(\begin{array}{ccc} \frac{9}{5}u_j & \mathcal {E}e_j & -\frac{\mathcal {E}}{T}u_j \\ \mathcal {E}e_j^\mathrm {Tr} & \rho \mathcal {E}u_j\mathbb {I} & 0 \\ -\frac{\mathcal {E}}{T}u_j & 0 & \frac{5\mathcal {E}^2}{6T^2}u_j \end{array}\right)\) 
and the matrix \(A_4^n\)  and column vector \(\mathcal {F}_n\)  can be written down easily. Here the symbol \(\mathbb {I}\)  denotes the \(3 \times 3\)  identity matrix and \(e^j,\,j=1,2,3\)  the standard (row) base vectors in \(\mathbb {R}^3\) , the symbol \((\cdot )^\mathrm {Tr}\)  means the transpose of row vectors.
Obviously, the matrix \(A_0^n\)  is positive definite. For the existence of local-in-time smooth solutions, we refer to the standard Friedrichs' theory (c.f. Chapter 2 in [1]}).
Now we devote ourselves to lifespan of the equations (REF ) and the estimate (REF ). It suffices to establish a priori energy estimates of (REF ). Taking \(\partial ^\beta \)  of (REF ) to get
\(\begin{aligned}A_0^n \partial _t \partial ^\beta U_n + \sum _{j=1}^{3}A_j^n \partial _j \partial ^\beta U_n + A_4^n \partial ^\beta U_n = & \partial ^\beta \mathcal {F}_n - \sum _{|\alpha |\leqslant |\beta |-1}\partial ^{\beta -\alpha }A_0^n \partial _t\partial ^\alpha U_n\\&\qquad - \sum _{j=1}^{3}\sum _{|\alpha |\leqslant |\beta |-1}\partial ^{\beta -\alpha }A_j^n \partial _j \partial ^\alpha U_n - \sum _{|\alpha |\leqslant |\beta |-1}\partial ^{\beta -\alpha }A_4^n \partial ^\alpha U_n.\end{aligned}\) 
Then taking inner product with \(\partial ^\beta U_n\)  in \(L^2(\mathrm {d}x)\)  and integrating over \([0,t]\) , we obtain
\(\begin{aligned}&\left(A_0^n \partial ^\beta U_n(t),\partial ^\beta U_n(t)\right)_{L^2(\mathrm {d}x)}-\left(A_0^n \partial ^\beta U_n(0),\partial ^\beta U_n(0)\right)_{L^2(\mathrm {d}x)}\\=&\int _{0}^{t}\left(\left[\partial _t A_0^n+\sum _{j=1}^{3}\partial _j A_j^n-(A_4^n+(A_4^n)^{Tr})\right]\partial ^\beta U_n(s),\partial ^\beta U_n(s)\right)_{L^2(\mathrm {d}x)}\mathrm {d}s+\int _{0}^{t}2\left(\partial ^\beta \mathcal {F}_n,\partial ^\beta U_n\right)_{L^2(\mathrm {d}x)}\mathrm {d}s\\&- \int _{0}^{t}2\sum _{|\alpha |\leqslant |\beta |-1}\left(\partial ^{\beta -\alpha }A_0^n \partial _t\partial ^\alpha U_n(s)+\sum _{j=1}^{3}\partial ^{\beta -\alpha }A_j^n \partial _j \partial ^\alpha U_n(s)+\partial ^{\beta -\alpha }A_4^n \partial ^\alpha U_n(s),\partial ^\beta U_n(s)\right)_{L^2(\mathrm {d}x)}\mathrm {d}s.\end{aligned}\) 
Since \(A_0^n\)  is positive definite, summing up over \(|\beta |\leqslant s_1\)  above, we arrive at
\(\left\Vert U_n(t)\right\Vert _{\mathcal {H}^{s_1}}^2 \leqslant \left\Vert U_n(0)\right\Vert _{\mathcal {H}^{s_1}}^2 + C(\mathbf {E}_{0,s_1+2}) \int _{0}^{t}\left\Vert U_n(s)\right\Vert _{\mathcal {H}^{s_1}}^2 \mathrm {d}s + C\int _{0}^{t}\left\Vert \mathcal {F}_n(s)\right\Vert _{\mathcal {H}^{s_1+2}}^2 \mathrm {d}s.\) 
A simple Grönwall's in equality implies that
\(\left\Vert U_n(t)\right\Vert _{\mathcal {H}^{s_1}}^2 \leqslant \left\lbrace \left\Vert U_n(0)\right\Vert _{\mathcal {H}^{s_1}}^2+ C\int _{0}^{t}\left\Vert \mathcal {F}_n(s)\right\Vert _{\mathcal {H}^{s_1+2}}^2 \mathrm {d}s\right\rbrace \left\lbrace 1+C(\mathbf {E}_{0,s_1+2})te^{C(\mathbf {E}_{0,s_1+2})t}\right\rbrace ,\) 
It follows that
\(\sup _{t\in (0,\tau )}\left\Vert U_n(t)\right\Vert _{\mathcal {H}^{s_1}}^2 \leqslant C(\tau ,\mathbf {E}_{0,s_1+2})\mathbb {E}_{n,s_1},\) 
which is exactly (REF ).
